Participation and demand levels for a joint project

Abstract

We examine a voluntary participation game in public good provision in which each agent has a demand level for the public good. The agent’s demand level is the minimum level of the public good from which she can receive a positive benefit. In this game, there exists a subgame perfect Nash equilibrium at which the (Pareto) efficient allocation is achieved. The voluntary participation game may also have a subgame perfect Nash equilibrium with underprovision of the public good. However, some subgame perfect Nash equilibrium with the efficient allocation satisfies strong perfection, introduced by Rubinstein (Int J Game Theory 9:1–12, 1980), and strong perfection is satisfied only by the subgame perfect Nash equilibrium with the efficient allocation. Furthermore, all payoffs at strong perfect equilibria belong to the core of the enterprise game. By these results, we conclude that in our case, the voluntary participation problem is not as serious as the earlier studies report. We also discuss the extensibility of these results.

Notes

Acknowledgments

I am indebted to the anonymous referees for their detailed comments and suggestions. I am grateful to Koichi Tadenuma, Koji Takamiya, Noriaki Matsushima, Motohiro Sato, Toshiji Miyakawa, Florian Morath, Ryo Arawatari, and Tomomi Miyazaki for their useful discussions and suggestions. I also thank the participants in the seminars at Shinshu University, Hitotsubashi University, Yokohama National University, Okayama University, Kyoto University, Hosei University, the 66th Congress of the International Institute of Public Finance, the 2010 autumn meeting of the Japanese Economic Association, the 67th Congress of the Japanese Institute of Public Finance, and the 65th European Meeting of the Econometric Society. This research was supported by a Grant-in-Aid for Scientific Research (21730156).

Appendix 1: Proofs of propositions

Proof of Proposition 1

(i)

Suppose that \(x_i(y)>0\) for some \(i\in P\) such that \(Y_i>{\psi }^P({\varvec{x_P}})\) and some \(y \in \bar{{\mathcal {Y}}}\). Then, \(i\) obtains the payoff \(-\sum _{z=1}^{t}x_i(\bar{y}^z)<0\). If she reduces her contribution, then she is made better off.

(ii)

Suppose that \(x_i(\bar{y}^l)>0\) for some \(i \in P \cap {\mathcal {N}}(0,{\psi }^P({\varvec{x_P}})]\) and some \(l \in \{1,\ldots ,t\}\) such that \(\bar{y}^l\in \bar{{\mathcal {Y}}}\) and \(\bar{y}^l>Y_i\). Note that since \(Y_i \ge \bar{y}^1\), \(l\ge 2\). Let \({\varvec{x_i^{\prime }}}=(x^{\prime }_i(y))_{y \in \bar{{\mathcal {Y}}}}\) be such that \(x_i^{\prime }(\bar{y}^l)=0\) and \(x_i^{\prime }(y)=x_i(y)\) for each \(y \in \bar{{\mathcal {Y}}}{\setminus } \{\bar{y}^l\}\). We first consider the case of \({\psi }^P({\varvec{x_P}})<\bar{y}^l\). In this case, by the definition of \({\psi }^P\), \({\psi }^P({\varvec{x_P}})={\psi }^P({\varvec{x_i^{\prime }}},{\varvec{x_{P{\setminus } \{i\}}}})\). Thus, \(\pi ^P_i({\varvec{x_P}})=\theta _i-\sum _{z=1}^t x_i(\bar{y}^z)<\theta _i-\sum _{z=1}^t x^{\prime }_i(\bar{y}^z)=\pi ^P_i({\varvec{x_i^{\prime }}},{\varvec{x_{P{\setminus } \{i\}}}})\), which implies that \({\varvec{x_P}}\) is not a Nash equilibrium. We also consider the case of \({\psi }^P({\varvec{x_P}}) \ge \bar{y}^l\). If \(i\) switches from \({\varvec{x_i}}\) to \({\varvec{x_i^{\prime }}}\), \(\bar{y}^l\) may not be provided. By the construction of \({\varvec{x_i^{\prime }}}\) and the definition of \({\psi }^P\), \({\psi }^P({\varvec{x_i^{\prime }}},{\varvec{x_{P{\setminus } \{i\}}}}) \ge \bar{y}^{l-1}\). Since \(\bar{y}^l> Y_i\) and \(\bar{{\mathcal {Y}}}\) is discrete, then \(\bar{y}^{l-1} \ge Y_i\). Then, \({\psi }^P({\varvec{x_i^{\prime }}},{\varvec{x_{P{\setminus } \{i\}}}})\ge Y_i\). Thus, if \(i\) switches from \({\varvec{x_i}}\) to \({\varvec{x_i^{\prime }}}\), then her demand level is fulfilled and her contribution declines.

(iii)

Suppose to the contrary that for some \(k \in \{1,\ldots ,t\}\) such that \(\bar{y}^k \in \bar{{\mathcal {Y}}}\) and \(\bar{y}^k \le {\psi }^P({\varvec{x_P}})\), \(\sum _{j \in P \cap {\mathcal {N}}(\bar{y}^{k-1},{\psi }^P({\varvec{x_P}})]}x_j(\bar{y}^k)>c(\bar{y}^k)-c(\bar{y}^{k-1})\) or \(\theta _i-\sum _{z=k}^t x_i(\bar{y}^z)<0\) for some \(i \in P \cap {\mathcal {N}}(\bar{y}^{k-1},{\psi }^P({\varvec{x_P}})]\). If the former holds, there is \(j \in P \cap {\mathcal {N}}(\bar{y}^{k-1},{\psi }^P({\varvec{x_P}})]\) such that \(x_j(\bar{y}^k)>0\). Agent \(j\) can reduce the marginal contribution from \(x_j(\bar{y}^k)\) slightly such that \({\psi }^P({\varvec{x_P}})\) is still provided and she is made better off. If the latter holds, \(\sum _{z=k}^t x_i(\bar{y}^z)>0\) because \(\theta _i>0\). If \(i\) sets her total contribution to zero, her payoff is at least 0. Thus, she is made better off. In any case, \({\varvec{x}}_P\) is not a Nash equilibrium.\(\square \)

Proof of Proposition 2

We first show (2.2).

(Necessity) Suppose to the contrary that a strong Nash equilibrium \({\varvec{x_P}} \in \mathbb {R}_+^{t|P|}\) does not satisfy GE for \(P\). Since \({\varvec{x_P}}\) is a Nash equilibrium and it satisfies the budget balance condition by (2), then \({\psi }^P({\varvec{x_P}}) \notin \arg \max _{y \in \{0\} \cup \bar{{\mathcal {Y}}}}\sum _{i \in P}B_i(y)-c(y)\). Let \(\bar{y} \equiv \max \ \arg \max _{y \in \{0\} \cup \bar{{\mathcal {Y}}}}\sum _{i \in P}B_i(y)-c(y)\). Then \(\sum _{j \in P \cap {\mathcal {N}}({\psi }^P({\varvec{x_P}}),\bar{y}]}\theta _i>c(\bar{y})-c({\psi }^P({\varvec{x_P}}))\). By (i) of Proposition 1, \(P \cap {\mathcal {N}}({\psi }^P({\varvec{x_P}}),\bar{y}]\) can coordinate their contributions in such a way that all of its members are made better off, which is a contradiction. Since \({\varvec{x_P}}\) is a Nash equilibrium, then by (i), (ii), and (3) in (iii) of Proposition 1, IR holds.

We finally show (2.1). A weakly strong Nash equilibrium is immune to all coalition deviations that make all coalition members strictly better off as well as unilateral deviations; hence, our strong Nash equilibrium is a weakly strong Nash equilibrium by definition. Branzei et al. (2005) show the existence of the core of \((N,W)\) and that for each core element, there exists a weakly strong Nash equilibrium whose equilibrium payoff is the core element. Thus, the weakly strong Nash equilibrium exists. We can show that weakly strong and strong Nash equilibria are equivalent in the contribution game as follows. By the sufficiency of (2.2), all Nash equilibria satisfying GE and IR for \(P\) are weakly strong Nash equilibria. In a similar way to the necessity in the above proof, we can show that all weakly strong Nash equilibria satisfy GE and IR for \(P\). \(\square \)

Proof of Proposition 3

Let \({\varvec{x_P}}\in \mathbb {R}_+^{t|P|}\) be a strong Nash equilibrium in the corresponding contribution game. If there exists \(i \in P \cap {\mathcal {N}}(0,{\psi }^P({\varvec{x_P}})]\) such that \(\sum _{y\in \bar{{\mathcal {Y}}}} x_i(y)=0\), then there is \(j \in P \cap {\mathcal {N}}(0,{\psi }^P({\varvec{x_P}})]{\setminus } \{i\}\) such that \(x_j(\bar{y})>0\) for some \(\bar{y} \le Y_i\). If we reduce the contribution of \(j\) slightly and add it to \(i\)’s contribution, then \(i\)’s contribution is positive. \(\square \)

Proof of Proposition 4

(Necessity) Suppose to the contrary that \(x_i^P>0\) and \(y^{P{\setminus } \{i \}} \ge Y_i\) for some \(i \in P\). By IR, \(y^P \ge Y_i\). If \(i\) participates, the payoff to \(i\) is \(\theta _i-x_i^P\). However, if \(i\) deviates and chooses not to participate, \(i\)’s payoff is \(\theta _i\). Thus, \(i\) is made better off, a contradiction. If \(y^{P\cup \{j\}}\ge Y_j>y^{P}\) and \(\theta _j - x_j^{P \cup \{i\}}>0\) for some \(j \notin P\), then \(j\) is better off participating in \(P\). Hence, for each \(i \notin P\), if \(y^{P\cup \{i\}}\ge Y_i>y^{P}\), then \(\theta _i - x_i^{P \cup \{i\}}\le 0\). By IR for \(P\cup \{i\}\), \(\theta _i - x_i^{P \cup \{i\}}= 0\).

We next prove that \(P^*\) satisfies IS: \(y^{P^* {\setminus } \{j\}}<Y_j\) for each \(j\in P^*\). Suppose to the contrary that there exists \(i\in P^*\) such that \(y^{P^* {\setminus } \{i\}} \ge Y_i\). Let \(l \in \{1,\ldots ,m\}\) be such that \(i \in P^l\). Let \(k \in \{1,\ldots ,m\}\) be such that \(y^k \in {\mathcal {Y}}\) and \(y^k = y^{P^* {\setminus } \{i\}}\). We first show \(m>k \ge l >l-1\). Since \(P^l \subseteq {\mathcal {N}}(y^{l-1},y^l]\), \(y^{l-1}<Y_i \le y^l\). Since \(y^{l-1}<Y_i\) and \(Y_i \le y^k\), then \(l-1<k\). Since \({\mathcal {Y}}\) is discrete and \(y^l\) and \(y^{l-1}\) lie next to each other, it is impossible that \(l> k > l-1\). Thus, we have \(k \ge l > l-1\). By the construction of \(P^*\), \(m>k\).8

Proof of Theorem 2

We first construct a strategy profile and then show that it is feasible (Lemma 1) and strongly perfect (Lemma 2). Let \(y^* \in \arg \max _{y \in \{0\} \cup \bar{{\mathcal {Y}}}} \sum _{j \in N}B_j(y)-c(y)\). We can take a subgame perfect Nash equilibrium \(s^{*}=(s^{*1},(\gamma ^{*}_i)_{i \in N}) \in {\mathcal {S}}\) on the path of which \(P^*\) provides \(y^*\) as the method before Theorem 1. For each \(i \in P^*\), denote \({\varvec{g_i}}=(g_i^z)_{z=1}^t \equiv \left( \gamma _i^{*z}\left( P^*\right) \right) _{z=1}^{t}\). Based on \(s^*\), we construct another strategy profile \(s=(s^1,(\gamma _i)_{i \in N}) \in {\mathcal {S}}\) in which every set of participants \(P \subseteq N\) produces the same level of the public good as that at \(s^*\), but the cost-shares may be different. Define \(P(s) = P^*\). Define \((\gamma _i(P))_{i \in P}\) for each \(P \subseteq N\) as follows:

Properties (c.1) and (c.2) are the same as (i)–(iii) in Proposition 1. In this section, in addition to these conditions, we require (c.3). This property means that if a set of participants switches from \(P^*\) to \(P\) such that \(P \cap P^* \ne \emptyset \) but not \(P^* \subseteq P\), then, the common members between these sets whose demand level is met contribute at least the level before the switch. This property plays an important role in proving that \(s\) is a strong perfect equilibrium. We first show that \(s\) is feasible.

Lemma 1

Proof

It is clear that \((\gamma _i)_{i \in N}\) can be constructed in a way that satisfies (a) and (b). Thus, we focus on each \(P \subseteq N\) such that \(P\cap P^* \ne \emptyset \) but not \(P^* \subseteq P\).

Let \(k \in \{1,\ldots ,l-1\}\). Suppose that \(\gamma _i^z(P)\) is defined in a way that satisfies (c.1)–(c.3) for each \(i \in P\) and each \(z \in \{k+1,\ldots ,l \}\). We now construct \(\gamma _i^k(P)\) for each \(i \in P\). Define \(\gamma _i^k(P)=0\) for each \(i \in P \cap {\mathcal {N}}(0,\bar{y}^{k-1}]\).

Claim 1 provides a sufficient condition for \(\gamma _i^k(P)\) for each \(i \in P\) to satisfy (c.3) for \(k\).

Note also that \((\gamma _i(P))_{i \in P}\) satisfies GE and IR for each \(P \subseteq N\), and hence, by Proposition 2, \((\gamma _i)_{i \in N}\) assigns a strong Nash equilibrium for each set of participants. Clearly, \(s\) is one of the efficient subgame perfect Nash equilibria. We now show that \(s\) is a strong perfect equilibrium.

Proof of Claim 2

At \(s\), each \(i \in {\mathcal {N}}(0,\bar{y}^{*}] {\setminus } P(s)\) receives the payoff \(\theta _i\), which is the greatest payoff that \(i\) can obtain. Thus, even if \(i\) joins in the deviation of \(D\), then she neither contributes nor is made better off.\(\square \)

Claim 3

For each \(i \in D\cap {\mathcal {N}}(y^{*},\bar{y}^t]\), if \(i\) is made better off by the deviation of \(D\), then there exists another agent \(j \in D\) that is made worse off by the deviation.

Proof of Claim 3

Suppose that \(i \in D \cap {\mathcal {N}}(y^{*},\bar{y}^t]\) is made better off by the deviation. Before the deviation, \(i\)’s payoff is zero. Hence, if \(i\) is made better off, then \(P(s^{\prime })\) provides at least \(Y_{i}\) units of the public good; otherwise, her payoff is not positive. We assume that \(P(s^{\prime })\) provides \(Y_i\) units of the public good at \(s^{\prime }\).9 If \(P(s^{\prime })\) provides \(Y_i\) units of the public good, it contributes \(c(Y_i)\) in total. Suppose that at \(s\), \(P(s^{\prime })\) provides \(\bar{y}^l\) for some \(l \in \{0,1,\ldots ,t\}\) such that \(\bar{y}^l \in \arg \max _{y \in \{0\} \cup \bar{{\mathcal {Y}}}} \sum _{i \in P(s^{\prime })}B_i(y)-c(y)\). Since \(y^*\) is the maximal efficient demand level supported at \((\gamma _{j})_{j\in N}\), \(\bar{y}^l \le y^*\). Thus, \(\bar{y}^l \le y^* < Y_i\). Since \(s_{-D}=s^{\prime }_{-D}\), \(D\cap P(s^{\prime })\) must contribute \(c(\bar{y}^l)-\sum _{j \in P(s^{\prime })\cap {\mathcal {N}}(0,\bar{y}^l]{\setminus } D}\sum _{z=1}^l \gamma _j^z(P(s^{\prime }))\) to the provision of \(\bar{y}^l\) units of the public good and \(c(Y_i)-c(\bar{y}^l)\) to increase a public good from \(\bar{y}^l\) to \(Y_i\) units.10 Hence,

By Claim 3, the payoff to each \(j \in {\mathcal {N}}(y^*,\bar{y}^t] \cap D\) is the same at \(s\) and \(s^{\prime }\); otherwise, the deviation by \(D\) is not profitable. Since the payoff to \(j\) at \(s\) is zero, \(j\)’s payoff is also zero at \(s'\), which means that the total contribution from \(j\) is zero at \(s^{\prime }\).

Claim 4

If an agent in \(D \cap {\mathcal {N}}(0,y^*]\cap P(s)\) is made better off, then there is another agent in \(D\) that is made worse off.

Proof of Claim 4

Suppose that \(i \in D \cap {\mathcal {N}}(0,y^*]\cap P(s)\) is made better off. Agent \(i\) receives the payoff \(\theta _i-\sum _{z=1}^t \gamma _i^z(P) \ge 0\) at \(s\). Thus, if \(i\) is made better off, her demand level is fulfilled at \(s^{\prime }\): otherwise, her payoff at \(s^{\prime }\) is at most zero. Since by Claims 2 and 3, no agents outside \(P(s) \cap {\mathcal {N}}(0,y^*]\) contribute a positive fee at \(s^{\prime }\) and their behaviors have no bearing on whether \(Y_i\) is fulfilled, it is sufficient to focus on the strategies of agents in \(P(s) \cap {\mathcal {N}}(0,y^*]\). By the change of strategies of agents in \(P(s) \cap {\mathcal {N}}(0,y^*]\), \(P(s^{\prime }) \subseteq P(s)\).

First, consider the case of \(s_i^{\prime }=1\): agent \(i\) continues to participate after the deviation. For \(i\) to be made better off, \(i\)’s total contribution at \(s^{\prime }\) is less than that at \(s\). However, by the construction of \((\gamma _i)_{i \in N}\), if \(i \in D\) reduces her contribution, then \(i\)’s demand level is not fulfilled. Then, agent \(k \in D{\setminus } \{i\}\) must increase her contribution to fulfill \(i\)’s demand level, which implies that \(k\) is made worse off.